Problem: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $18.5$ years; the standard deviation is $4.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $31.1$ years.
Solution: $18.5$ $14.3$ $22.7$ $10.1$ $26.9$ $5.9$ $31.1$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $18.5$ years. We know the standard deviation is $4.2$ years, so one standard deviation below the mean is $14.3$ years and one standard deviation above the mean is $22.7$ years. Two standard deviations below the mean is $10.1$ years and two standard deviations above the mean is $26.9$ years. Three standard deviations below the mean is $5.9$ years and three standard deviations above the mean is $31.1$ years. We are interested in the probability of a tiger living less than $31.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $5.9$ years and the other half $({0.15\%})$ will live longer than $31.1$ years. The probability of a particular tiger living less than $31.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.